What is the extreme f(x)=x^{1/2}-x^{3/2} ?

The extreme f(x)=x^{1/2}-x^{3/2} is Minimum(0,0),Maximum(1/3 ,(1/3)^{1/2}-(1/3)^{3/2})

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extreme f(x)=x^{1/2}-x^{3/2}

extreme points

f (x) = x^{\frac{1}{2}} - x^{\frac{3}{2}}

Minimum (0, 0), Maximum (\frac{1}{3}, (\frac{1}{3})^{\frac{1}{2}} - (\frac{1}{3})^{\frac{3}{2}})

Solution steps

f^{'} (x) = \frac{1}{2 x ^{\frac{1}{2}}} - \frac{3 x ^{\frac{1}{2}}}{2}

Find intervals:

Increasing

: 0 < x < \frac{1}{3},

Decreasing

: \frac{1}{3} < x < \infty

Plug

x = 0

into

x^{\frac{1}{2}} - x^{\frac{3}{2}} : 0

Minimum (0, 0)

Plug

x = \frac{1}{3}

into

x^{\frac{1}{2}} - x^{\frac{3}{2}} : (\frac{1}{3})^{\frac{1}{2}} - (\frac{1}{3})^{\frac{3}{2}}

Maximum (\frac{1}{3}, (\frac{1}{3})^{\frac{1}{2}} - (\frac{1}{3})^{\frac{3}{2}})

Minimum (0, 0), Maximum (\frac{1}{3}, (\frac{1}{3})^{\frac{1}{2}} - (\frac{1}{3})^{\frac{3}{2}})

f (χ) = χ^{2} e^{- 4 χ^{2}}

f (x) = \frac{y - 800}{- 2}

(x - 1) e^{\frac{1}{x}}

f (x) = x^{4} + x^{2} (y - 2) + 3 (y - 1)^{2}

f (x, y) = 2 e^{x} ln (1 + y)

What is the extreme f(x)=x^{1/2}-x^{3/2} ?
The extreme f(x)=x^{1/2}-x^{3/2} is Minimum(0,0),Maximum(1/3 ,(1/3)^{1/2}-(1/3)^{3/2})